INTEGERS 13 (2013)
#A60
SOME WEIGHTED SUMS OF POWERS OF FIBONACCI
POLYNOMIALS
Claudio Pita Ruiz
Department of Mathematics, Universidad Panamericana, Mexico City, Mexico
cpita@up.edu.mx
Received: 10/8/12, Revised: 4/6/13, Accepted: 8/10/13, Published: 9/26/13
Abstract
We obtain closed formulas for some weighted Psums of powers of bivariate
q
n
k
Fibonacci and Lucas polynomials of the form
(x, y) Ftsn+l
(x, y) and
n=0
Pq
n
k
(x, y) Ltsn+l (x, y), in the cases k = 1, 2, 3, 4, and some specific values
n=0
of the parameters t and l. We express these sums as linear combinations of the
Fibopolynomials q+m
tk Fs (x,y) , m = 1, 2, . . . , tk.
1. Introduction
The sequence of generalized bivariate Fibonacci polynomials Gn (x, y) is defined by
the second-order recurrence Gn+2 (x, y) = xGn+1 (x, y) + yGn (x, y), with arbitrary
initial conditions G0 (x, y) and G1 (x, y). When G0 (x, y) = 0 and G1 (x, y) = 1
we have the bivariate Fibonacci polynomials Fn (x, y), and when G0 (x, y) = 2 and
G1 (x, y) = x we have the bivariate Lucas polynomials Ln (x, y). (What we will use
about these polynomials is contained in reference [2].) The corresponding extensions
n
to negative indices are given by F n (x, y) = ( y) Fn (x, y) and L n (x, y) =
n
( y) Ln (x, y), n 2 N, respectively. In the case y = 1, we have the Fibonacci and
Lucas polynomials (in the variable x), Fn (x, 1) and Ln (x, 1), denoted simply as
Fn (x) and Ln (x). By setting x = 1 in these polynomials, we obtain the numerical
sequences Fn (1) and Ln (1), denoted as Fn and Ln , corresponding to the Fibonacci
sequence Fn = (0, 1, 1, 2, 3, 5, . . .) and the Lucas sequence Ln = (2, 1, 3, 4, 7, 11, . . .).
In a recent work [7] we showed that the sequence Gktsn+µ (x, y), where t, s, k 2 N
and µ 2 Z are given parameters, can be written as a linear combination of the
i
(s-)Fibopolynomials n+tk
, i = 0, 1, . . . , tk, according to
tk
Fs (x,y)
Gktsn+µ
s+1
(x, y) = ( 1)
tk X
i
X
( 1)
i=0 j=0
⇥Gkts(i
j)+µ
(x, y) y
✓
◆
tk + 1
j
Fs (x,y)
✓
◆
1)
n + tk i
.
tk
Fs (x,y)
(sj+2(s+1))(j+1)
2
sj(j
2
(1)
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Recall that for integers n, p 0, we have n0 Fs (x,y) = nn Fs (x,y) = 1 and
✓ ◆
Fsn (x, y) Fs(n 1) (x, y) · · · Fs(n p+1) (x, y)
n
=
,
0 < p < n.
p Fs (x,y)
Fs (x, y) F2s (x, y) · · · Fps (x, y)
If n or p are negative, or p > n, we have np
= 0. It is known that np
Fs (x,y)
Fs (x,y)
are indeed polynomials in x and y. When x = y = 1 we have the s-Fibonomials
n
n
p Fs (see [6]), and when x = y = s = 1, we have the (usual) Fibonomials p F ,
introduced and studied by Hoggatt [3] in 1967.
If (x, y) is a (non-zero) given real function of the real variables x and y, we see
from (1) that
q
X
n
(x, y) Gktsn+µ (x, y)
n=0
tk X
i
X
s+1
= ( 1)
( 1)
i=0 j=0
⇥Gkts(i
j)+µ
(x, y) y
✓
◆
tk + 1
j
Fs (x,y)
✓
◆
q
X
n + tk i
n
(x, y)
.
tk
Fs (x,y)
n=0
(sj+2(s+1))(j+1)
2
sj(j 1)
2
(2)
Expression (2) can be written as
q
X
n=0
n
(x, y) Gktsn+µ (x, y)
✓
◆
tk + 1
= ( 1)
( 1)
j
Fs (x,y)
m=1 i=0 j=0
✓
◆
sj(j 1)
q+m
k
i tk+q+m
2
⇥Gts(i j)+µ (x, y) y
(x, y)
tk
Fs (x,y)
✓
◆
tk
i
XX
(sj+2(s+1))(j+1)
tk + 1
s+1
tk
2
+ ( 1)
(x, y)
( 1)
j
Fs (x,y)
i=0 j=0
✓
◆
q
X
sj(j 1)
n
i
n
⇥Gkts(i j)+µ (x, y) y 2
(x, y)
(x, y)
.
tk Fs (x,y)
n=0
s+1
tk tk
i
X
Xm X
(sj+2(s+1))(j+1)
2
(3)
This simple observation derived from (1) allows us to obtain closed formulas, in
Pq
terms of Fibopolynomials, for the weighted sums n=0 n (x, y) Gktsn+µ (x, y), as
the following proposition states (with a straightforward proof using (3)).
Proposition 1. If z =
(x, y) is a root of
✓
◆
tk X
i
X
(sj+2(s+1))(j+1)
tk + 1
2
( 1)
Gkts(i
j
Fs (x,y)
i=0 j=0
j)+µ
(x, y) y
sj(j 1)
2
z i = 0,
(4)
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Pq
then the weighted sum of k-th powers n=0 n (x, y) Gktsn+µ (x, y) can be expressed
as a linear combination of the s-Fibopolynomials q+m
tk Fs (x,y) , m = 1, 2, . . . , tk,
according to
q
X
n q
(x, y) Gktsn+µ (x, y)
n=0
s+1
= ( 1)
tk tk
i
X
Xm X
( 1)
m=1 i=0 j=0
⇥Gkts(i
j)+µ
(x, y) y
sj(j 1)
2
✓
◆
tk + 1
j
Fs (x,y)
✓
◆
q+m
tk+m
(x, y)
.
tk
Fs (x,y)
(sj+2(s+1))(j+1)
2
i
Moreover, suppose expression (5) is valid for some weight function
z = (x, y) is a root of (4).
(5)
(x, y). Then
Let us consider the simplest case t = k = 1. Equation (4) is in this case
0 =
1 X
i
X
( 1)
(sj+2(s+1))(j+1)
2
i=0 j=0
= Gµ (x, y) + (Gs+µ (x, y)
If Gs+µ (x, y)
✓ ◆
2
Gs(i
j Fs (x,y)
j)+µ
(x, y) y
sj(j 1)
2
zi
(6)
Ls (x, y) Gµ (x, y)) z.
Ls (x, y) Gµ (x, y) 6= 0, we have from (6) the weight
(x, y) =
Gµ (x, y)
.
Ls (x, y) Gµ (x, y) Gs+µ (x, y)
(7)
Expression (5) for the corresponding weighted sum is in this case
q
X
n
(x, y) Gsn+µ (x, y) = Gµ (x, y)
q
(x, y)
n=0
✓
◆
q+1
.
1
Fs (x,y)
(8)
In the Fibonacci case we have, from (7) and (8), that if µ 6= s then
q ✓
X
n=0
Fµ (x, y)
s
( y) Fµ s (x, y)
◆n
q
Fsn+µ (x, y) =
Fµ (x, y)
Fs(q+1) (x, y) .
Fs (x, y)
(9)
Similarly, from (7) and (8), we have in the Lucas case that
q ✓
X
n=0
Lµ (x, y)
s
( y) Lµ s (x, y)
◆n
q
Lsn+µ (x, y) =
Lµ (x, y)
Fs(q+1) (x, y) .
Fs (x, y)
(10)
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Thus, if µ 6= 0, s we can combine (9) and (10) together as
◆n q
q ✓
X
1
Fµ (x, y)
Fsn+µ (x, y)
Fµ (x, y) n=0 ( y)s Fµ s (x, y)
◆n q
q ✓
X
1
Lµ (x, y)
=
Lsn+µ (x, y)
Lµ (x, y) n=0 ( y)s Lµ s (x, y)
=
1
Fs(q+1) (x, y) .
Fs (x, y)
(11)
1
We call attention to the fact that the term Fs (x,y)
Fs(q+1) (x, y) in (11) does not depend on the parameter µ. That is, in (11) we have infinitely many weighted sums of
1
q+1 bivariate Fibonacci and Lucas polynomials that are equal to Fs (x,y)
Fs(q+1) (x, y).
We can have a slight generalization of (11) if we substitute x by Lr (x, y) and y
r+1 r
by ( 1)
y , where r 2 Z. By using the equations
⇣
⌘
r+1 r
Fr (x, y) Fn Lr (x, y) , ( 1)
y
= Frn (x, y) ,
⇣
⌘
r+1 r
Ln Lr (x, y) , ( 1)
y
= Lrn (x, y) ,
we get for r 6= 0 (and µ 6= 0, s in the Fibonacci sums)
◆n q
q ✓
X
1
Frµ (x, y)
Fr(sn+µ) (x, y)
Frµ (x, y) n=0 ( y)rs Fr(µ s) (x, y)
◆n q
q ✓
X
1
Lrµ (x, y)
=
Lr(sn+µ) (x, y)
Lrµ (x, y) n=0 ( y)rs Lr(µ s) (x, y)
=
1
Frs(q+1) (x, y) .
Frs (x, y)
(12)
Some examples from (12) (corresponding to some specific values of µ) are the
following:
• With s = 1 we have
◆n q
q ✓
X
1
F2r (x, y)
Fr(n 2) (x, y)
F 2r (x, y) n=0 F3r (x, y)
◆n q
q ✓
X
1
1
=
Fr(n 1) (x, y)
F r (x, y) n=0 Lr (x, y)
◆n q
q ✓
X
1
Lr (x, y)
=
Fr(n+2) (x, y)
r
F2r (x, y) n=0
( y)
◆n q
q ✓
X
1
F3r (x, y)
=
Fr(n+3) (x, y)
F3r (x, y) n=0 ( y)r F2r (x, y)
(13)
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◆n q
q ✓
X
1
L2r (x, y)
Lr(n 2) (x, y)
L 2r (x, y) n=0 L3r (x, y)
◆n q
◆n q
q ✓
q ✓
X
1X
2
1
Lr (x, y)
=
Lrn (x, y) =
Lr(n+1) (x, y)
2 n=0 Lr (x, y)
Lr (x, y) n=0 2 ( y)r
◆n q
q ✓
X
Fr(q+1) (x, y)
1
Lr (x, y)
=
Lr(n 1) (x, y) =
.
L r (x, y) n=0 L2r (x, y)
Fr (x, y)
=
• With s = 2 we have
=
=
=
=
◆n q
q ✓
X
1
F3r (x, y)
Fr(2n 3) (x, y)
F 3r (x, y) n=0 F5r (x, y)
◆n q
q ✓
X
1
Fr (x, y)
Fr(2n 1) (x, y)
F r (x, y) n=0 F3r (x, y)
◆n q
q ✓
X
1
F3r (x, y)
Fr(2n+3) (x, y)
F3r (x, y) n=0 y 2r Fr (x, y)
◆n q
q ✓
X
1
L3r (x, y)
Lr(2n 3) (x, y)
L 3r (x, y) n=0 L5r (x, y)
◆n q
q ✓
X
1
Lr (x, y)
Lr(2n 1) (x, y)
L r (x, y) n=0 L3r (x, y)
(14)
q
q
( 1)rq X ( 1)rnLr(2n+1)(x, y) ( 1)(r+1)q X ( 1)(r+1)nFr(2n+1)(x, y)
=
=
Lr (x, y) n=0
Fr (x, y) n=0
y r(n q)
y r(n q)
◆n q
q ✓
X
F2r(q+1) (x, y)
1
L3r (x, y)
=
Lr(2n+3) (x, y) =
.
2r
L3r (x, y) n=0 y Lr (x, y)
F2r (x, y)
• With s = 3 we have
◆n q
q ✓
X
1
1
Fr(3n+1) (x, y)
Fr (x, y) n=0 ( y)r Lr (x, y)
◆n q
q ✓
X
1
Lr (x, y)
=
Fr(3n+2) (x, y)
F2r (x, y) n=0
y 2r
!n q
q
X
1
Lr (x, y) L2r (x, y)
=
Fr(3n+4) (x, y)
3r
F4r (x, y) n=0
( y)
◆n q
q ✓
r
( y) X
1
=
Fr(3n 1) (x, y)
Fr (x, y) n=0 Lr (x, y) L2r (x, y)
(15)
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◆n q
q ✓
X
1
Lr (x, y)
Lr(3n 1) (x, y)
L r (x, y) n=0 L4r (x, y)
◆n q
q ✓
X
1
Lr (x, y)
=
Lr(3n+1) (x, y)
Lr (x, y) n=0 ( y)r L2r (x, y)
◆n q
q ✓
X
1
L2r (x, y)
=
Lr(3n+2) (x, y)
L2r (x, y) n=0 y 2r Lr (x, y)
!n q
q
X
F3r(q+1) (x, y)
1
L4r (x, y)
=
Lr(3n+4) (x, y) =
.
3r
L4r (x, y) n=0 ( y) Lr (x, y)
F3r (x, y)
=
In particular we have the following numerical identities (from (13), (14) and (15)
with x = y = r = 1)
Fq+1
q
X
=
2q
n
Fn
n=0
q
X
2
=
1
n
( 2)
2 n=0
=
q
X
=
n q 1
2
q
=
=
n=0
q
X
q
X
=
=
=
=
1
q
X
n+q
( 1)
n=0
◆n q
q ✓
1X 3
3 n=0 4
q n
( 2)
Fn+2
Ln
=
2
=
q
X
(16)
q
X
3n
n=0
1
q
Ln
1
Ln+1 .
n=0
n q 1
2
F2n
3
q
X
2q
1
4 n=0
L2n+1 =
F3n+1 =
q
X
4n
n
4
11
F2n
◆n
q 1
1
F2n+1
(17)
n=0
q
L2n
3
=
q
X
4q
n
L2n
1
n=0
L2n+3 .
q
X
n q
( 1)
F3n+2 =
n=0
1
3n
n=0
n=0
q
X
n=0
n=0
q
X
q
X
=
n=0
q ✓
X
F2n+3 =
n+q
( 1)
q
n=0
1
F3(q+1)
2
Fn
Fn+3 =
Ln =
q ✓ ◆n
1X 2
2 n=0 5
q
X
q
X
n=0
n=0
F2(q+1)
=
F
q 3n
n q 1
3
1
=
L3n+2
q
X
1
L
n q 3n
7
n=0
q
1X
n
=
( 7)
7 n=0
q
q
1X
n q
( 3)
F3n+4
(18)
3 n=0
◆n q
q ✓
X
1
=
L3n+1
1
3
n=0
L3n+4 .
Of course, some of the numerical results shown in (16), (17) and (18) are wellPq
known identities. For example, in [1]: (i) formula n=0 2n q 1 Ln = Fq+1 of (16) is
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INTEGERS: 13 (2013)
Pq
identity 236, (ii) formula n=0 F2n+1 = F2(q+1) of (17) is essentially identity 2, and
Pq
(iii) formula n=0 F3n+1 = 12 F3(q+1) in (18) is essentially identity 24 (corrected).
In the remaining sections of this work we consider only the case y = 1 of Proposition 1. The corresponding weight function (x, 1) will be denoted as (x).
2. Weighted Sums of Cubes
In this section we will obtain expressions for some weighted sums of cubes of Fibonacci polynomials. More specifically, we will consider the Fibonacci case of Proposition 1 when k = 3, µ = 0 and t = 1.
In this case, equation (4) is (after some simplifications)
s
Fs3 (x) z z 2 + 2Ls (x) z + ( 1)
Thus, we have two weights, namely
q
s
L2s (x) ( 1) ,
1 (x) = Ls (x) +
2
(x) =
= 0.
Ls (x)
(19)
q
L2s (x)
s
( 1) . (20)
The corresponding weighted sum (5) can be written as
!
✓
◆
✓
◆
q
X
q+1
q+2
s+1
n q
3
3
(x) Fsn (x) = Fs (x) ( 1)
(x)
+
, (21)
3
3
Fs (x)
Fs (x)
n=0
where (x) is any of the weights (20).
In the case x = 1, we have in particular the following numerical identities (obtained by setting s = 1 and s = 2 in (21) with the corresponding weights (20))
q ⇣
X
n=0
q ⇣
X
n=0
1±
p ⌘n
3±2 2
p ⌘n
2
q
3
F2n
=
q
Fn3 =
p ⌘
Fq+1 Fq ⇣⇣
1 ± 2 Fq
2
1
p ⌘
F2(q+1) F2q ⇣ ⇣
3 ± 2 2 F2(q
24
⌘
+ Fq+2 .
1)
(22)
⌘
+ F2(q+2) . (23)
We will show now some di↵erent versions of (21), involving Chebyshev polynomials of the first kind Tn (x), or of the second kind Un (x). The results are the
following.
Proposition 2. (a) If s is even, we have the following weighted sums of cubes of
Fibonacci polynomials, valid for any integer l 0
q
X
3
Tn+l ( Ls (x)) Fsn
(x)
n=1
=
Fs3
(x)
(24)
!
✓
◆
✓
◆
q+1
q+2
Tq+l+1 ( Ls (x))
+ Tq+l ( Ls (x))
.
3
3
Fs (x)
Fs (x)
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q
X
Un+l
n=1
1
3
( Ls (x)) Fsn
(x)
(25)
!
✓
◆
q+2
.
1 ( Ls (x))
3
Fs (x)
✓
◆
q+1
Uq+l ( Ls (x))
+ Uq+l
3
Fs (x)
= Fs3 (x)
(b) If s is odd, we have the following weighted sums of cubes of Fibonacci polynomials (where i2 = 1), valid for any integer l 0
q
X
n
3
( i) Tn+l ( iLs (x)) Fsn
(x)
(26)
n=1
q+1
= Fs3 (x) ( i)
Tq+l+1 ( iLs (x))
✓
◆
q+1
3
Fs (x)
!
✓
◆
q+2
+ ( i) Tq+l ( iLs (x))
3
Fs (x)
q
q
X
n 1
( i)
Un+l
1
3
( iLs (x)) Fsn
(x)
n=1
.
(27)
✓
◆
q+1
=
(x) ( i) Uq+l ( iLs (x))
3
Fs (x)
!
✓
◆
q+2
q 1
+ ( i)
Uq+l 1 ( iLs (x))
.
3
Fs (x)
Fs3
q
Proof. Recall that Chebyshev polynomials of the first kind Tn (x) can be calculated
as
⌘n ⇣
⌘n ⌘
p
p
1 ⇣⇣
Tn (x) =
x + x2 1 + x
x2 1
,
(28)
2
and that Chebyshev polynomials of the second kind Un (x) can be calculated as
✓⇣
⌘n+1 ⇣
⌘n+1 ◆
p
p
1
Un (x) = p
x + x2 1
x
x2 1
.
(29)
2 x2 1
We can write (21) as
q
X
n+l
3
(x)Fsn
(x)
(30)
n=0
=
Fs3
s+1
(x) ( 1)
q+l+1
✓
◆
q+1
(x)
+
3
Fs (x)
q+l
!
✓
◆
q+2
(x)
,
3
Fs (x)
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where l is a non-negative integer. If s is even, the weights (20) are
p
p
L2s (x) 1 ,
L2s (x)
1 (x) = Ls (x) +
2 (x) = Ls (x)
1.
(31)
Thus, (24) follows from (28), (30) and (31). Similarly, (25) follows from (29),
(30) and (31).
p
On the other hand,
if
s
is
odd
the
weights
(20)
are
(x)
=
L
(x)+
L2s (x) + 1,
1
s
p
2
Ls (x) + 1. These weights can be written as
2 (x) = Ls (x)
✓
◆
q
2
(x)
=
i
iL
(x)
+
(
iL
(x))
1
,
(32)
1
s
s
✓
◆
q
2
(x)
=
i
iL
(x)
(
iL
(x))
1
.
2
s
s
From (28) and (32) we see that
n
( i) Tn ( iLs (x)) =
1
(
2
n
1
(x) +
n
2
(x)) ,
(33)
and from (29) and (32) we see that
n+1
( i)
Un ( iLs (x)) =
1
p
2
2i Ls (x) + 1
n+1
1
(x)
n+1
2
(x) .
(34)
Thus, (26) follows from (30), (32) and (33). Similarly, (27) follows from (30),
(32) and (34).
We note that the integer parameter l
0 gives us, in (24) and (25), infinitely
many weighted sums of cubes of Fibonacci polynomials for each even s, and in (26)
and (27) gives us infinitely many weighted sums of cubes of Fibonacci polynomials
for each odd s.
We show some numerical examples from formulas (24) to (27). We set x = 1 and
q = 5 in them.
1
If s = 2, the sequence (Tn ( L2 ))n=1 involved in (24) is
1
(Tn ( L2 ))n=1 = ( 3, 17, 99, 577, 3363, 19601, 114243, 665857, . . .) .
(See [5, A001541] for the unsigned version.) Thus, we have the following weighted
sums of cubes of Fibonacci numbers (corresponding to l = 0, 1)
3F23 + 17F43
17F23
99F63 + 577F83
99F43 + 577F63
3
3363F10
=
3
3363F83 + 19601F10
=
F10 F12
( 19601F8 3363F14 ) ,
L2 F6
F10 F12
(114243F8 + 19601F14 ) .
L2 F6
1
Similarly, the sequence (Un ( L2 ))n=1 involved in (25) is
1
(Un ( L2 ))n=1 = ( 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, . . .) .
10
INTEGERS: 13 (2013)
(See [5, A001109] for the unsigned version.) Thus, we have (for l = 1, 2)
6F23 + 35F43
35F23
204F63 + 1189F83
204F43 + 1189F63
3
6930F10
=
F10 F12
( 40391F8
L2 F6
3
6930F83 + 40391F10
=
6930F14 ) ,
F10 F12
(235416F8 + 40391F14 ) .
L2 F6
1
n
If s = 1, the sequence (( i) Tn ( iL1 ))n=1 involved in (26) is
1
n
(( i) Tn ( iL1 ))n=1 = ( 1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, . . .) .
(See [5, A001333] for the unsigned version.) Then, we have the weighted sums
(for l = 0, 1)
F13 + 3F23
3F13
7F33 + 17F43
7F23 + 17F33
41F53
=
41F43 + 99F53
=
F6 F5
(99F4 41F7 ) ,
2
F6 F5
( 239F4 + 99F7 ) .
2
1
n
Similarly, if s = 3, the sequence (( i) Un ( iL3 ))n=1 involved in (27) is
1
n
(( i) Un ( iL3 ))n=1
= ( 8, 65, 528, 4289, 34840, 283009, 2298912, 18674305, . . .) .
(See [5, A041025] for the unsigned version.) Then, we have the weighted sums
(for l = 1, 2)
8F33 + 65F63
3
528F93 + 4289F12
3
34840F15
=
65F33
528F63 + 4289F93
F32 F18 F15
(283009F12
F6 F9
34840F21 ) ,
3
3
34840F12
+ 283009F15
=
F32 F18 F15
( 2298912F12 + 283009F21 ) .
F6 F9
3. Other Weighted Sums
In this section we will obtain expressions for some other weighted sums of Fibonacci
and Lucas polynomials. More specifically, we will set µ = 0 in Proposition 1 and
consider the following cases: (i) k = t = 2 and G = F or G = L; (ii) k = 4, t = 1,
G = F.
11
INTEGERS: 13 (2013)
Let us begin with the case (i). When G = F , equation (4) is
4 X
i
X
0 =
( 1)
(sj+2(s+1))(j+1)
2
i=0 j=0
s+1
= ( 1)
✓ ◆
5
F2
j Fs (x) 2s(i
2
F2s
(x, y) z (z + 1) z 2
j)
(x) z i
(35)
s
( 1) L2s (x) z + 1 ,
and with G = L is
4 X
i
X
0 =
( 1)
(sj+2(s+1))(j+1)
2
i=0 j=0
s+1
= ( 1)
L22s (x) z 2
✓ ◆
5
L2
j Fs (x) 2s(i
j)
(3L4s (x) + 2) z + 4
(x) z i
(36)
s
z2
( 1) L2s (x) z + 1 .
Observe that
s
z2
s
s
( 1) L2s (x) z + 1 = z ( 1) ↵2s (x) z ( 1) 2s (x) ,
p
p
where ↵ (x) = 12 x + x2 + 4 and (x) = 12 x
x2 + 4 .
s
2
Since z
( 1) L2s (x) z + 1 is a factor of the right-hand sides of (35) and (36),
we have for both, the Fibonacci and the Lucas cases, the following weights
1
s
(x) = ( 1) ↵2s (x) ,
2
s
(x) = ( 1)
2s
(x) .
(37)
In the Fibonacci case we have in addition the weight (x) = 1 (from the factor
z + 1 of the right-hand side of (35)). In the Lucas case we have in addition the
following weights (from the factor L22s (x) z 2 (3L4s (x) + 2) z + 4 of the right-hand
side of (36))
0
1
s✓
◆2
2
3L
(x)
+
2
3L
(x)
+
2
4s
4s
@
+
1A ,
3 (x) =
L2s (x)
4L2s (x)
4L2s (x)
0
1
s✓
◆2
2
3L4s (x) + 2
@ 3L4s (x) + 2
1A .
(38)
4 (x) =
L2s (x)
4L2s (x)
4L2s (x)
In the Fibonacci case the corresponding sum (5) is
q
X
1
2 (x)
F2s
n=0
n q
2
(x) F2sn
✓
◆
✓
◆
q+1
q+3
(x) =
(x)
+
(39)
4
4
Fs (x)
Fs (x)
✓
◆✓
◆
q+2
s+1 L3s (x)
1
+
(x) + ( 1)
,
Ls (x)
4
Fs (x)
12
INTEGERS: 13 (2013)
(where
(5) is
q
X
(x) is any of the weights (37) or
n q
(x) L22sn (x) =
L22s (x) (x)
n=0
(x) =
1). In the Lucas case the sum
✓
◆
✓
◆
q+1
q+4
+4
4
4
Fs (x)
Fs (x)
s
+ ( 1) L3s (x) L3s (x) (x)
+ 4
1
✓
◆
q+2
4
Fs (x)
✓
◆
q
+
3
s
2 ( 1) )
,
4
Fs (x)
L22s (x)
L2s (x) (3L2s (x)
(x)
(40)
2
(x)
(where (x) is any of the weights (37) or (38)).
For the weight (x) = 1 of the Fibonacci case, we have from (39) the following
alternating sum of squares of Fibonacci polynomials
q
X
n+q
( 1)
n=0
=
2
F2s
2
F2sn
(x)
(41)
!
✓
◆
✓
◆
✓
◆
q+1
q+2
q+3
s+1
(x)
+ ( 1)
L2s (x)
+
.
4
4
4
Fs (x)
Fs (x)
Fs (x)
p
When x = s = 1, the weights (37) are 3±2 5 . In this case we have the following
numerical formulas for weighted sums of squares of Fibonacci and Lucas numbers
in terms of Fibonomials:
q ⇣
X
n=0
q ⇣
X
n=0
p ⌘n q
3± 5
2
p ⌘n q
3± 5
2
2
F2n
=
p
3± 5 q+1
2
4 F
L22n
=
p
9(3± 5) q+1
2
4 F
+
17
+
p
±2 5
p
5± 5 q+2
2
4 F
+
q+3
4 F,
p
51±23 5 q+2
2
4 F
q+3
4 F
+4
q+4
4 F.
By using the weights (37) in (39) and (40), we can obtain expressions for weighted
sums of squares of Fibonacci and Lucas polynomials, in which the weight functions
are in turn certain Fibonacci or Lucas polynomials. This is the content of the
following proposition.
Proposition 3. For l 2 Z we have the following weighted sums of squares of
Fibonacci and Lucas polynomials
INTEGERS: 13 (2013)
13
(a)
q
X
1
s(n+q+1)+1
2
( 1)
F2s(n+l) (x) F2sn
(x)
2 (x)
F2s
n=0
✓
◆
✓
◆
q+1
q+2
= F2s(q+l+1) (x)
+ Fs (x)Ls(2q+2l+1) (x)
4
4
Fs (x)
Fs (x)
✓
◆
q+3
s+1
+ ( 1) F2s(q+l) (x)
.
4
Fs (x)
(42)
(b)
q
X
1
s(n+q+1)+1
2
( 1)
L2s(n+l) (x) F2sn
(x)
(43)
2 (x)
F2s
n=0
✓
◆
✓
◆
q+1
q+2
2
= L2s(q+l+1) (x)
+ x + 4 Fs (x) Fs(2q+2l+1) (x)
4
4
Fs (x)
Fs (x)
✓
◆
q+3
s+1
+ ( 1)
L2s(q+l) (x)
.
4
Fs (x)
(c)
q
X
s(n+q)
( 1)
F2s(n+l) (x) L22sn (x)
(44)
n=0
✓
◆
✓
◆
q+1
q+4
=
(x) ( 1)
F2s(q+l+1) (x)
+ 4F2s(q+l) (x)
4
4
Fs (x)
Fs (x)
✓
◆
q+2
+ L3s (x) L3s (x) F2s(q+l+1) (x) L22s (x) F2s(q+l+2) (x)
4
Fs (x)
✓
◆
q+3
s
s
2
+ 4 ( 1) F2s(q+l 1) (x) Ls (x) (3L2s (x) 2 ( 1) ) F2s(q+l) (x)
.
4
Fs (x)
s+1
L22s
(d)
q
X
s(n+q)
( 1)
L2s(n+l) (x) L22sn (x)
n=0
(45)
✓
◆
✓
◆
q+1
q+4
+ 4L2s(q+l) (x)
4
4
Fs (x)
Fs (x)
✓
◆
q
+
2
+ L3s (x) L3s (x) L2s(q+l+1) (x) L22s (x) L2s(q+l+2) (x)
4
Fs (x)
✓
◆
q+3
s
s
+ 4 ( 1) L2s(q+l 1) (x) L2s (x) (3L2s (x) 2 ( 1) ) L2s(q+l) (x)
.
4
Fs (x)
s+1
= ( 1)
L22s (x) L2s(q+l+1) (x)
14
INTEGERS: 13 (2013)
Proof. First write the sum (39) as
q
X
1
2 (x)
F2s
n=0
n+l
2
(x)F2sn
(x)
(46)
q+l+1
=
+
✓
(x)
q+l 1
✓
◆
q+1
+
4
Fs (x)
s+1
(x) + ( 1)
✓
◆
q+3
4
Fs (x)
◆✓
◆
L3s (x) q+l
q+2
(x)
,
Ls (x)
4
Fs (x)
q+l
(x)
where l 2 Z. Substitute the weights (37) in (46), take the di↵erence of the resulting
expressions, multiply both sides of this di↵erence by x2 + 4
r
formula Fr (x) = px12 +4 (↵r (x)
(x)), to obtain that
1
2
, and use the Binet’s
q
X
1
2
( 1)s(n+q+1)+1 F2s(n+l) (x) F2sn
(x)
2 (x)
F2s
n=0
✓
◆
✓
◆
q+1
q+3
s+1
= F2s(q+l+1) (x)
+ ( 1)
F2s(q+l) (x)
4
4
Fs (x)
Fs (x)
✓
◆✓
◆
L3s (x)
q+2
+
F2s(q+l) (x) F2s(q+l+1) (x)
.
Ls (x)
4
Fs (x)
Finally, use the identity
L3s (x)
F2s(q+l) (x)
Ls (x)
F2s(q+l
1)
(x) = Fs (x) Ls(2q+2l+1) (x) ,
to obtain (42). Similarly, if we substitute the weights (37) in (46), then take the sum
of the resulting expressions, then use the Binet’s formula Lr (x) = ↵r (x) + r (x),
and then use the identity
L3s (x)
L2s(q+l) (x)
Ls (x)
L2s(q+l
1)
(x) = x2 + 4 Fs (x) Fs(2q+2l+1) (x) ,
we obtain (43).
In a similar fashion, if we begin now with (40), written as
q
X
n+l
(x)L22sn (x)
(47)
n=0
=
L22s (x)
q+l+1
s
(x)
✓
◆
q+1
+4
4
Fs (x)
+ ( 1) L3s (x) L3s (x)
+ 4
q+l 1
(x)
q+l+1
(x)
L2s (x) (3L2s (x)
✓
◆
q+4
4
Fs (x)
✓
◆
q+2
2
q+l+2
L2s (x)
(x)
4
Fs (x)
✓
◆
q+3
s
2 ( 1) ) q+l (x)
,
4
Fs (x)
q+l
(x)
15
INTEGERS: 13 (2013)
we see that (44) and (45) are obtained by using the weights (37) in (47), together
with Binet’s formulas.
Observe that the case l = 0 of (42) and (45) gives us sums of cubes of Fibonacci and Lucas polynomials. More precisely, if s is even we have the following
(unweighted) sums of cubes
q
X
1
3
F2sn
(x)
2 (x)
F2s
n=0
=
q
X
(48)
✓
◆
q+1
4
Fs (x)
✓
◆
q+3
+ F2sq (x)
.
4
Fs (x)
F2s(q+1) (x)
Fs (x) Ls(2q+1) (x)
✓
◆
q+2
4
Fs (x)
✓
◆
✓
◆
q+1
q+4
+ 4L2sq (x)
(49)
4
4
Fs (x)
Fs (x)
✓
◆
q+2
+ L3s (x) L3s (x) L2s(q+1) (x) L22s (x) L2s(q+2) (x)
4
Fs (x)
✓
◆
q+3
+ 4L2s(q 1) (x) Ls (x) (3L3s (x) + Ls (x)) L2sq (x)
.
4
Fs (x)
L32sn (x) =
n=0
L22s (x) L2s(q+1) (x)
If s is odd, we have the following alternating sums of cubes
q
X
1
n+q 3
( 1)
F2sn (x)
2 (x)
F2s
n=0
(50)
q
X
(51)
✓
◆
✓
◆
q+1
q+2
= F2s(q+1) (x)
+ Fs (x) Ls(2q+1) (x)
4
4
Fs (x)
Fs (x)
✓
◆
q+3
+ F2sq (x)
.
4
Fs (x)
n=0
n+q
( 1)
L32sn (x)
✓
◆
✓
◆
q+1
q+4
+ 4L2sq (x)
4
4
Fs (x)
Fs (x)
✓
◆
q+2
3
2
+ Ls (x) L3s (x) L2s(q+1) (x) L2s (x) L2s(q+2) (x)
4
Fs (x)
✓
◆
q+3
4L2s(q 1) (x) + Ls (x) (3L3s (x) Ls (x)) L2sq (x)
.
4
Fs (x)
= L22s (x) L2s(q+1) (x)
16
INTEGERS: 13 (2013)
For the remaining weights (38) of the Lucas case, we have in particular the
following numerical identity (case s = x = 1)
q ⇣
X
n=0
⌘n q
p
23± 385
18
L22n =
p
23± 385 q+1
2
4 F
p
61±3 385 q+2
2
4 F
p
1± 385 q+3
2
4 F
+4
q+4
4 F.
In fact, we can proceed in a similar fashion of the proof of Proposition 2 (using
now (38) and (40)) to obtain that, if
s
(x) =
3L4s (x) + 2
,
4L2s (x)
we have the following weighted sum of squares of Lucas polynomials, valid for
integers l 0,
◆n q
q ✓
X
2
Tn+l ( s (x)) L22sn (x)
(52)
L
(x)
2s
n=0
✓
◆
q+1
= 2L2s (x) Tq+l+1 ( s (x))
4
Fs (x)
✓
◆✓
◆
3
q+2
s 2Ls (x) L3s (x)
+ ( 1)
Tq+l+1 ( s (x)) 4Tq+l+2 ( s (x))
L2s (x)
4
Fs (x)
✓
◆
q+3
s
+ 2L2s (x) Tq+l 1 ( s (x)) L2s (x) (3L2s (x) 2 ( 1) ) Tq+l ( s (x))
4
Fs (x)
✓
◆
q+4
+ 4Tq+l ( s (x))
.
4
Fs (x)
For example, if x = 1 and s = 2, the sequence Tn
143
28
1
n=0
involved in (52) is
1
Tn 143
28
n=0
20057 2840123 402206417 56958853523 8066283596057 1142314618945643
= 1, 143
,
,
,...
28
392 , 5488 ,
76832 ,
1075648 ,
15059072
210827008
For q = 4 and l = 0, 1, we have the weighted sums
2
7
=
4
3 143 2
2 20057 2
2
2
7
28 L4 + 7
392 L8
56958853523 5
400320800329 6
76832
1 F2 +
76832
2 F2
L20 +
1 2840123 2
2
402206417 2
7
5488 L12 +
76832 L16
7
68220633199
402206417 8
76832
3 F2 +
19208
4 F2 ,
+
and
2
7
4 143 2
28 L0
+
+
=
+
3 20057 2
392 L4
2
1 402206417 2
2
2840123 2
2
56958853523 2
7
5488 L8 + 7
76832 L12 +
1075648 L16
8066283596057 5
56691820586491 6
9661131494701 7
1075648
1 F2 +
1075648
2 F2
1075648
3 F2
56958853523 8
.
268912
4 F2
2
7
.
17
INTEGERS: 13 (2013)
Finally, let us consider the case (ii) mentioned at the beginning of this section
(Fibonacci case of Proposition 1, with µ = 0, k = 4 and t = 1). In this case,
equation (4) is
4 X
i
X
0 =
( 1)
(sj+2(s+1))(j+1)
2
i=0 j=0
s+1
= ( 1)
(x) =
3
(x) =
j)
(x) z i
s
(53)
s
Fs4 (x) z (z + 1) z 2 + ( 1) (3L2s (x) + 4 ( 1) ) z + 1 .
Thus we have the weights
2
✓ ◆
5
F4
j Fs (x) s(i
1 and
s
✓
s
s ◆2
3L2s (x) + 4 ( 1)
3L2s (x) + 4 ( 1)
+
s
2
2 ( 1)
s
✓
s
s ◆2
3L2s (x) + 4 ( 1)
3L2s (x) + 4 ( 1)
s
2
2 ( 1)
1
(x) =
1,
1.
(54)
The corresponding weighted sum (5) (of fourth powers of Fibonacci polynomials)
is in this case
q
1 X
Fs4 (x) n=0
n q
4
(x)Fsn
(x)
=
(55)
✓
◆
✓
◆
q+1
q+3
(x)
+
4
4
Fs (x)
Fs (x)
s
(3 ( 1) L2s (x) + 5) (x) +
2
(x)
✓
◆
q+2
.
4
Fs (x)
With the weight 1 (x) = 1 we obtain from (55) the following alternating sum
of fourth powers of Fibonacci polynomials
q
X
n
4
( 1) Fsn
(x)
n=0
q
= ( 1)
Fs4
(56)
!
✓
◆
✓
◆
✓
◆
q+1
q+2
q+3
s
(x)
+ (3 ( 1) L2s (x) + 4)
+
.
4
4
4
Fs (x)
Fs (x)
Fs (x)
(The case x = 1 of (56), after some transformations, is contained in [4].) By
setting s = 1 in the weights (54), we have in particular the numerical identity from
(55)
q ⇣
p ⌘n q
p
p
X
q+3
5± 21
3± 21 q+2
Fn4 = 5±2 21 q+1
(57)
2
4 F
2
4 F +
4 F.
n=0
18
INTEGERS: 13 (2013)
However, following the ideas of the proof of Proposition 2, we can see from (54)
and (55) that for any integer l 0 we have
q
1 X
4
Tn+l (⇥s (x)) Fsn
(x)
(58)
Fs4 (x) n=1
✓
◆
✓
◆
q+1
q+3
= Tq+l+1 (⇥s (x))
+ Tq+l (⇥s (x))
4
4
Fs (x)
Fs (x)
✓
◆
q+2
s
((3 ( 1) L2s (x) + 5) Tq+l+1 (⇥s (x)) + Tq+l+2 (⇥s (x)))
,
4
Fs (x)
where
s
⇥s (x) =
3L2s (x) + 4 ( 1)
.
s
2 ( 1)
For example, if x = 1 and s = 2, we have ⇥2 (1) =
25 1
Tn
involved in (58) is
2
n=1
Tn
=
25 1
2
n=1
25 623
388127
,
,
2
2 , 7775,
2
(See [5, A090733] for 2Tn
sums
23 4
2 F2
+
623 4
2 F4
4
7775F6
9687625
, 120901249,
2
25
2
+
4
7775F4
+
388127 4
F6
2
=
and the sequence
6035374825 150642568127
,
,...
2
2
.
.) With q = 5 and l = 0, 1 we have the weighted
388127 4
F8
2
9687625 4
F10
2
6
251490123 7
120901249 2
2
3 F2
F2
=
623 4
2 F2
25
2 ,
4
9687625 4
F8 + 120901249F10
2
6035374825 6
6277177323 7
2
2 F2 +
2
3 F2
9687625 8
2
4 F2 ,
+ 120901249
8
4 F2 .
Acknowledgements I thank the referee for a careful reading of the first version of
this work, and all comments and suggestions that certainly contributed to present
a more readable version of the article.
References
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arXiv:math/0406323v1
formulae
for
bivariate
Fibonacci
and
Lucas
polynomials,
[3] V. E. Hoggatt, Jr. Fibonacci numbers and generalized binomial coefficients, Fibonacci Quart.
5 (1967), 383–400.
INTEGERS: 13 (2013)
19
[4] R. S. Melham, Alternating sums of fourth powers of Fibonacci and Lucas numbers, Fibonacci
Quart. 38 (2000), 254–259.
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[7] C. Pita, On bivariate s-Fibopolynomials, arXiv:1203.6055v1